The Soc for People of 'GRDCT2008' Mk VI Watch

This discussion is closed.
Champagne Supernova
Badges: 20
#6581
Report 9 years ago
#6581
(Original post by jaz_jaz)
Wait, I thought acknowledgment meant they've got it, its not a rejection right now but they're considering it.
Cuz I got an email from York the day after saying your application has been recieved and acknowledged :confused:
Yeah, an acknowledgement is when they say 'just to let you know, we've got your application, and will give you more info later (i.e. offer/interview/rejection).
ArchedEdge
Badges: 15
Rep:
?
#6582
Report 9 years ago
#6582
(Original post by GHOSH-5)
 \dfrac{\mathrm{d}}{\mathrm{d}x} (\ln x(x+1)) = \dfrac{\mathrm{d}}{\mathrm{d}x} (\ln x + \ln (x+1)) = \dfrac{1}{x} + \dfrac{1}{x+1}

Omg that was ridiculously easy wasn't it?

I'm such a muppet at maths
0
John Locke
Badges: 11
Rep:
?
#6583
Report 9 years ago
#6583
(Original post by raceforthefishman)
How do I find out when the oxford interviews for biochemistry are???
http://www.ox.ac.uk/admissions/under...timetable.html

14th-15th Dec for all and 14th-16th for some apparently
0
raceforthefishman
Badges: 2
Rep:
?
#6584
Report 9 years ago
#6584
(Original post by GHOSH-5)
Is this:  \ln \dfrac{1}{3x+1} ?

Separate indeed! :smile:
Yes it was that!!

so Id get ln1-ln(3x+1)

ln1 would disappear magically
so then
-3/(3x+1)
:awesome:

*sorry about the lack of latex*

(Original post by John Locke)
14th-15th Dec for all and 14th-16th for some apparently
Thanks!
0
Unbounded
Badges: 12
Rep:
?
#6585
Report 9 years ago
#6585
(Original post by raceforthefishman)
Yes it was that!!

so Id get ln1-ln(3x+1)

ln1 would disappear magically
so then
-3/(3x+1)
:awesome:
All correct. :smile:
0
jaz_jaz
Badges: 0
Rep:
?
#6586
Report 9 years ago
#6586
OHHHHHHHHHHHHHHHH
I have 4/5 acknowledgements then
Erghh, lack of tubes and weather = postponed til a later date.
0
ArchedEdge
Badges: 15
Rep:
?
#6587
Report 9 years ago
#6587
Ghosh, help me!

How does one integrate 2sintcos^2t ?
0
Champagne Supernova
Badges: 20
#6588
Report 9 years ago
#6588
(Original post by ArchedEdge)
Ghosh, help me!

How does one integrate 2sintcos^2t ?
He's not online at the mo'
ArchedEdge
Badges: 15
Rep:
?
#6589
Report 9 years ago
#6589
(Original post by Champagne Supernova)
He's not online at the mo'
Damn that boy!
He'll do it sooner or later anyways
and I'm in no rush
0
Unbounded
Badges: 12
Rep:
?
#6590
Report 9 years ago
#6590
(Original post by ArchedEdge)
Ghosh, help me!

How does one integrate 2sintcos^2t ?
Make the orgasmic substitution of  u = \cos t .
Spoiler:
Show
 I = \displaystyle\int 2\sin t \cos^2 t \ \mathrm{d}t

 u = \cos t \implies \mathrm{d}u = -\sin t \mathrm{d}t \implies \mathrm{d}t = \dfrac{\mathrm{d}u}{-\sin t}

Shoving everything we already have in our integral:

 I = \displaystyle\int 2\sin t \ u^2 \ \dfrac{\mathrm{d}u}{-\sin t} = \displaystyle\int -2u^2 \ \mathrm{d}u

I expect some rep in return for helping you :awesome:
0
raceforthefishman
Badges: 2
Rep:
?
#6591
Report 9 years ago
#6591
(Original post by GHOSH-5)
Make the orgasmic substitution of  u = \cos t .
Spoiler:
Show
 I = \displaystyle\int 2\sin t \cos^2 t \ \mathrm{d}t

 u = \cos t \implies \mathrm{d}u = -\sin t \mathrm{d}t \implies \mathrm{d}t = \dfrac{\mathrm{d}u}{-\sin t}

Shoving everything we already have in our integral:

 I = \displaystyle\int 2\sin t \ u^2 \ \dfrac{\mathrm{d}u}{-\sin t} = \displaystyle\int -2u^2 \ \mathrm{d}u

I expect some rep in return for helping you :awesome:
Say what???
I'll look forward to this topic :sigh:
Is it in c4??????
0
overclocked
Badges: 12
Rep:
?
#6592
Report 9 years ago
#6592
:zomg:
bloody core 3.
0
Champagne Supernova
Badges: 20
#6593
Report 9 years ago
#6593
Great...got that to look forward to. :indiff:
Unbounded
Badges: 12
Rep:
?
#6594
Report 9 years ago
#6594
(Original post by raceforthefishman)
Say what???
I'll look forward to this topic :sigh:
Is it in c4??????
It is C4 - integration by substitution. It's 'sort of' the integral analogue of chain rule; ArchedEdge's integral is in fact a very clear example of 'inverse chain rule', but integration by substitution is not always the opposite of chain rule.
0
jaz_jaz
Badges: 0
Rep:
?
#6595
Report 9 years ago
#6595
Oh joy. As if C3 and FP1 weren't bad enough, FP2 and C4 to come :shoot:
0
raceforthefishman
Badges: 2
Rep:
?
#6596
Report 9 years ago
#6596
(Original post by GHOSH-5)
It is C4 - integration by substitution. It's 'sort of' the integral analogue of chain rule; ArchedEdge's integral is in fact a very clear example of 'inverse chain rule', but integration by substitution is not always the opposite of chain rule.
^^ :dry:

I don't know whether I remembered to say thank you for letting me use your maths brain...so if I didn't, thank you!
0
Unbounded
Badges: 12
Rep:
?
#6597
Report 9 years ago
#6597
(Original post by raceforthefishman)
^^ :dry:

I don't know whether I remembered to say thank you for letting me use your maths brain...so if I didn't, thank you!
You're welcome

edit: Cheers for being a rep-slave ArchedEdge :awesome:
0
jaz_jaz
Badges: 0
Rep:
?
#6598
Report 9 years ago
#6598
Ghoshhh, would you be able to explain the method to find the domain and range of functions.
I've asked literally every person I know and you're my last hope! :o:
0
Champagne Supernova
Badges: 20
#6599
Report 9 years ago
#6599
Southampton acknowledgement.
Unbounded
Badges: 12
Rep:
?
#6600
Report 9 years ago
#6600
(Original post by Champagne Supernova)
Southampton acknowledgement.
:woo:

Good news! St John's have said that if I'm called for interview, it'll be on the morning of 12th December, so I can just get a later flight to my ski trip
0
X
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Did you get less than your required grades and still get into university?

Yes (52)
28.42%
No - I got the required grades (108)
59.02%
No - I missed the required grades and didn't get in (23)
12.57%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise